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I am so stuck on this question and was wondering if you could help me out. How do you solve 2^3x+1 = 3^x-2 ? If you can help me it would be greatly appreciated.
Hi Laurel,
2^(3x + 1) = 3^(x - 2)
(2^3x).(2^1) = (3^x)(3^-2)
2(2^3x) = (3^x)/9
2(2^3)^x = (3^x)/9
2(8^x) = (3^x)/9
18 = 3^x / 8^x
18 = (3/8)^x
Taking logarithms of both sides
log 18 = log (3/8)^x
log 18 = x.log (3/8)
x = log 18 / log (3/8)
Regards
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Ahmed Salami
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I can provide good answers to questions dealing in almost all of mathematics especially from A`Level downwards. I can as well help a good deal in Physics with most emphasis directed towards mechanics.
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